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Gelneren [198K]

3 years ago

5

Question is in the pic

2 answers:

mihalych1998 [28]3 years ago

8 0

The answer is C in the picture.

Anton [14]3 years ago

4 0

Answer: The answer is A

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An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4

Maslowich

Answer:

Average speed = 10,000 m/s

Explanation:

Given the following data;

Distance = 2m

Time = 0.0002secs

To find the average speed;

Average speed = distance/time

Average speed = 2/0.0002

Average speed = 10,000 m/s

Therefore, the average speed of the

electron is 10,000 meters per seconds.

7 0

3 years ago

he calculation of kinetic energy uses the formula, KE = 1 2 mv2. Identify the unit that is required for mass and velocity, respe

Umnica [9.8K]

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Velocity is measured in ms^-1
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3 years ago

4. Why do you think there needs to be a specific code of ethics for health-related information on the Internet?

alexandr1967 [171]

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3 years ago

Read 2 more answers

Which of these objects would need the largest force to create an acceleration of 1 m/s2?

babymother [125]

I think the answer is D the 30 kg sofa

7 0

4 years ago

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

7 0

4 years ago